A gas occupies 4.00 L at 2.00 atm pressure. At what pressure would the volume be 6.00 L if the temperature remains constant?
2 answers:
Answer:
The final pressure would be 1.33 atm
Explanation:
<u>Given:</u>
Initial volume of the gas, V1 = 4.00 L
Initial pressure, P1 = 2.00 atm
Final volume, V2 = 6.00 L
<u>To determine:</u>
The final pressure P2 of the gas
<u>Explanation:</u>
Based on the ideal gas equation:
where: n = number of moles,
R = gas constant and T = temperature
At constant Temperature the gas law reduces to,
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Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.
Explanation : Given,
Moles of = 2.0 moles
Molar mass of = 32 g/mole
Now we have to calculate the moles of
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react to give 3 mole of
So, 2.0 moles of react to give
moles of
Now we have to calculate the mass of
Therefore, the mass of oxygen gas produced is, 96 grams.
Now we have to determine the pressure exerted by the gas against the container walls.
Using ideal gas equation:
where,
P = pressure of oxygen gas = ?
V = volume of oxygen gas
T = temperature of oxygen gas =
R = gas constant = 0.0821 L.atm/mole.K
w = mass of oxygen gas
= density of oxygen gas = 1.429 g/L
M = molar mass of oxygen gas = 32 g/mole
Now put all the given values in the ideal gas equation, we get:
Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.